Chapter 1 - Limits and Their Properties - 1.5 Exercises - Page 88: 8

$\lim\limits_{x\to4^-}f(x)=-\infty.$ $\lim\limits_{x\to4^+}f(x)=-\infty.$

$\lim\limits_{x\to4^-}f(x)=\lim\limits_{x\to4^-}\dfrac{-1}{(x-4)^2}=\dfrac{-1}{(4^--4)^2}=\dfrac{-1}{(0^-)^2}=\dfrac{-1}{0^+}=-\infty.$ $\lim\limits_{x\to4^+}f(x)=\lim\limits_{x\to4^+}\dfrac{-1}{(x-4)^2}=\dfrac{-1}{(4^+-4)^2}=\dfrac{-1}{(0^+)^2}=-\infty.$